Algebraic Curves∗

Thus in order to find all points of inflection of an algebraic curve we first determine the points
where the curvature is zero by finding the simultaneous solutions of the equations
f(x,y) = 0,
f 2 y f xx − 2f x f y f xy + f 2 x f yy = 0,
and then find those solutions which are non-singular points. Next, we determine whether the
curvature changes sign as we move along the curve past the point of zero curvature. If it does, the
point is an inflection. In practice we can check on the change of sign if, for example, the curve can
be parametrized near the point by x, and
f 2 y f xx − 2f x f y f xy + f 2 x f yy
becomes a function of x after substituting from f(x,y) = 0. Such a local parameterization by x
exists by Theorem 1 provided that the tangent at the point is not parallel to the y-axis.
Example 3. Calculate the curvature of the hyperbola
at the point (2, 1).
Let f = x 2 − 3y 2 − 1. We have
Therefore
κ =
x 2 − 3y 2 = 1
f x = 2x,
f y = −6y,
f xx = 2,
f xy = 0,
f yy = −6.
± (−6y)2 · 2 + (2x) 2 · (−6)
(4x 2 + 36y 2 ) 3 2
= ± 9y2 − 3x 2
.
(x 2 + 9y 2 ) 3 2
In particular,
κ(2, 1) = ± 3
( √ 13) 3 .
Example 4. Find the points of inflection of the acnodal cubic
f(x, y) = y 2 + x 2 − x 3 = 0.
We have
f x = 2x − 3x 2 ,
f y = 2y,
f xx = 2 − 6x,
f xy = 0,
f yy = 2.
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